Question: A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank $t$ minutes after the drain is opened: $V(t)=3000(1-0.05t)^2$ What is the instantaneous rate of change of the volume after $10$ minutes? Choose 1 answer: Choose 1 answer: (Choice A) A $-150$ liters per minute (Choice B) B $-150$ minutes per liter (Choice C) C $750$ liters per minute (Choice D) D $750$ minutes per liter
Solution: Understanding the problem The function that represents the instantaneous rate of change of $V(t)$ is its derivative, $V'(t)$. Therefore, the instantaneous rate of change of the volume $10$ minutes after the drain is opened is $V'(10)$. Let's find $V'(t)$ and evaluate it at $t=10$. Finding $V'(t)$ $V'(t)=15t-300$ Finding $V'(10)$ $\begin{aligned} V'(10)&=15(10)-300 \\\\ &=-150 \end{aligned}$ Interpreting units $V(t)$ is the volume of water in ${\text{liters}}$ after $t$ ${\text{minutes}}$. Therefore, we measure its rate of change in ${\text{liters}}$ per ${\text{minute}}$. In conclusion, the instantaneous rate of change of the volume after $10$ minutes is $-150$ liters per minute. The rate of change is negative because the amount of remaining water is decreasing.